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A 2-wire DC distributor cable 800 m long is loaded with 1 A/m. Resistance of each conductor is 0.05 Ω/km. Calculate the maximum voltage drop if the distributor is fed from both ends with equal voltages of 220 V.
- a)2 V
- b)8 V
- c)16 V
- d)4 V
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
A 2-wire DC distributor cable 800 m long is loaded with 1 A/m. Resista...
Concept:
Total current supplied by the distributor, I = 800 × 1 = 800A
Total resistance of the distributor , R = 2 × 0.05 × 0.8 = 0.08 ohm
For a uniformly loaded DC-distributed wire fed from both sides with equal voltages, the Vmin occurs at mid-point( x = l/2).
So the maximum voltage drop is at the mid-point:
Calculation:
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Community Answer
A 2-wire DC distributor cable 800 m long is loaded with 1 A/m. Resista...
To calculate the maximum voltage drop in the 2-wire DC distributor cable, we need to consider the resistance of the cable and the current flowing through it.
Given:
Length of the cable (L) = 800 m
Current per unit length (I) = 1 A/m
Resistance per unit length (R) = 0.05 Ω/km
Voltage at each end of the cable (V) = 220 V
1. Calculate the total resistance of the cable:
The resistance of the cable can be calculated using the formula:
Resistance (R) = Resistance per unit length (R') * Length of the cable (L)
In this case, R' = 0.05 Ω/km and L = 800 m.
So, R = 0.05 Ω/km * 800 m = 40 Ω
2. Calculate the total current flowing through the cable:
The total current flowing through the cable can be calculated using the formula:
Total current (I_total) = Current per unit length (I) * Length of the cable (L)
In this case, I = 1 A/m and L = 800 m.
So, I_total = 1 A/m * 800 m = 800 A
3. Calculate the voltage drop across the cable:
The voltage drop across the cable can be calculated using Ohm's Law:
Voltage drop (V_drop) = Resistance (R) * Total current (I_total)
In this case, R = 40 Ω and I_total = 800 A.
So, V_drop = 40 Ω * 800 A = 32000 V
4. Determine the maximum voltage drop:
Since the cable is fed from both ends with equal voltages, the maximum voltage drop will occur when the cable is fully loaded. In this case, the maximum voltage drop will be half of the total voltage drop.
Maximum voltage drop = V_drop / 2 = 32000 V / 2 = 16000 V
Therefore, the maximum voltage drop in the 2-wire DC distributor cable is 16000 V.
The correct answer is option C) 16 V.
Given:
Length of the cable (L) = 800 m
Current per unit length (I) = 1 A/m
Resistance per unit length (R) = 0.05 Ω/km
Voltage at each end of the cable (V) = 220 V
1. Calculate the total resistance of the cable:
The resistance of the cable can be calculated using the formula:
Resistance (R) = Resistance per unit length (R') * Length of the cable (L)
In this case, R' = 0.05 Ω/km and L = 800 m.
So, R = 0.05 Ω/km * 800 m = 40 Ω
2. Calculate the total current flowing through the cable:
The total current flowing through the cable can be calculated using the formula:
Total current (I_total) = Current per unit length (I) * Length of the cable (L)
In this case, I = 1 A/m and L = 800 m.
So, I_total = 1 A/m * 800 m = 800 A
3. Calculate the voltage drop across the cable:
The voltage drop across the cable can be calculated using Ohm's Law:
Voltage drop (V_drop) = Resistance (R) * Total current (I_total)
In this case, R = 40 Ω and I_total = 800 A.
So, V_drop = 40 Ω * 800 A = 32000 V
4. Determine the maximum voltage drop:
Since the cable is fed from both ends with equal voltages, the maximum voltage drop will occur when the cable is fully loaded. In this case, the maximum voltage drop will be half of the total voltage drop.
Maximum voltage drop = V_drop / 2 = 32000 V / 2 = 16000 V
Therefore, the maximum voltage drop in the 2-wire DC distributor cable is 16000 V.
The correct answer is option C) 16 V.
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